Problem: Simplify the following expression and state the condition under which the simplification is valid. You can assume that $q \neq 0$. $y = \dfrac{3q}{36q + 81} \times \dfrac{7(4q + 9)}{6q} $
Solution: When multiplying fractions, we multiply the numerators and the denominators. $y = \dfrac{ 3q \times 7(4q + 9) } { (36q + 81) \times 6q } $ $ y = \dfrac {3q \times 7(4q + 9)} {6q \times 9(4q + 9)} $ $ y = \dfrac{21q(4q + 9)}{54q(4q + 9)} $ We can cancel the $4q + 9$ so long as $4q + 9 \neq 0$ Therefore $q \neq -\dfrac{9}{4}$ $y = \dfrac{21q \cancel{(4q + 9})}{54q \cancel{(4q + 9)}} = \dfrac{21q}{54q} = \dfrac{7}{18} $